Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 54



Work Step by Step

Here, we have $a_n= a_1r^{(n-1)}$ for the Geometric series. We are told that $a_5=40.5$ and $r=-0.25$ Here, we have $a_2=(a_1) \times (-0.25)^{(5-1)} \implies 40.5=(a_1) \times (-0.25)^4$ or, $a_1=10368$ Now, $a_n=a_1 \times r^{n-1}$ Hence, $a_n=10368(-0.25)^{n-1}$
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